REDOX 3. Oxidation & Reduction Organic Preparations compound formation reactions in organic chemistry oxidation states in organic molecules

Redox reactions in Organic Chemistry A summary of some redox reactions used in organic synthesis is given below. Further details for selected reactions are given below the summary tables. Most of the reactions described are found in one or other of UK based GCE-AS-A2 or IB courses. The application of oxidation states to organic molecules can be tricky, but, (i) use of half-cell equations usually gets round the problem, and (ii) hopefully the oxidation state exemplars in the last section will help illuminate the situation if you are interested, but this knowledge is not required at this level? Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so. The concept of functional group level is dealt with in appendix 1.
8. Summary of some ORGANIC SYNTHESIS REDUCTION REACTIONS Guide notes: YES/NO denotes whether reaction possible. Lithium tetrahydridoaluminate(III), LiAlH₄ (lithium aluminium hydride), is a more powerful reducing agent than sodium tetrahydridoborate(III), NaBH₄ (sodium borohydride) and accounts for the NO/YES differences in columns (a) and (b). R = H, alkyl or aryl for aldehydes, carboxylic acids and nitriles, R = alkyl or aryl for ketones. Click on 8.1, 8.2, 8.3, 8.4 and 8.5 in the table for more details on (a), (b), (c) or (d) reaction reagents/conditions and the chemical outcome. Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so.
homologous series change on reduction	molecular structure change	(a) NaBH₄ in water (lab method)	(b) LiAlH₄ in e.g. ether (lab method)	(c) reflux with Sn/conc. HCl_(aq) (lab method)	(d) Ni/H₂ at 150^oC (industry) Pd and Pt are also used as catalysts
8.1 alkene ==> alkane	>C=C< ==> -CH-CH-	NO	NO	NO	YES, used
8.2 aldehyde/ketone ==> primary/secondary alcohol	RCHO ==> RCH₂OH and R₂C=O ==> R₂CHOH	YES	YES	can be reduced with Na/C₂H₅OH or Zn/CH₃COOH mixtures	YES, used?
8.3 carboxylic acid ==> primary aliphatic alcohol	RCOOH ==> RCH₂OH	NO	YES	NO	YES, used?
8.4 nitrile ==> primary aliphatic amine	RCN ==> RCH₂NH₂	NO	YES	NO	YES, used?
8.5 nitro-aromatic ==> primary aromatic amine	e.g. C₆H₅NO₂ ==> C₆H₅NH₂	NO	YES	YES	YES
Some further details of the organic reductions tabulated above 8.1 Reduction of alkenes to alkanes 8.1(a)/(b) Alkenes cannot be reduced by metal hydride complexes (see 8.2/8.3a-b) because the 'attacking' species is nucleophilic e.g. the negative (e^- pair donating) BH₄^- or AlH₄^- ions are repelled too strongly to allow 'disruption' of the electron rich C=C double bond in alkenes. Alkenes tend to react with electron pair seeking electrophilic reagents like H^δ+-Br^δ-. 8.1(c) Reduction reactions by metal/acid occur on the surface of the metal and involve electron transfer from the metal to the organic compound (see 8.2(c) equations). Presumably the electron rich C=C bond is not a strong enough electron acceptor as is the case with nucleophiles, but acts in the opposite way as an electron donor to electron deficient electrophiles. 8.1(d) e.g. propene ==> propane: CH₃CH=CH₂ + H₂ == Ni catalyst ==> CH₃CH₂CH₃ This reaction is initiated on the nickel catalyst surface which lowers the activation energy to help break the H-H bonds and 'half' of the C=C bond. (see diagram and explanation of the hydrogenation mechanism). The catalysed hydrogenation of alkenes is a very important industrial reaction used to convert natural polyunsaturated oils into low melting solid more saturated fats like margarine. 8.2 Reduction of aldehydes to primary alcohols and ketones to secondary alcohols These reactions are essentially the reduction of the carbony1 group >C=O to >CHOH. 8.2(a) Using sodium tetrahydrioborate(III), NaBH₄ (sodium borohydride) The reaction can be carried out in water. The reduction mechanism is very complicated, but can be considered in a simplistic way as involving the donation of a hydride ion to the aldehyde/ketone. An outline of the nucleophilic addition mechanism is given on the Organic Mechanisms Part III page, but the simple equation will suffice here. aldehyde: RCHO + 2[H] ==> RCH₂OH (R = H, alkyl or aryl) ketone: R₂C=O + 2[H] ==> R₂CHOH (R = alkyl or aryl) 8.2(b) Using lithium tetrahydridoaluminate(III), LiAlH₄ (lithium tetrahydride) LiAlH₄ is a more powerful reducing agent than NaBH₄ and reacts violently with water (and reacts with ethanol too?, so the reaction must be carried out in an inert solvent like ethoxyethane ('ether'). The initial product is hydrolysed by dil. sulphuric acid. The simplified equations above in 8.2(a) apply. 8.2(c) Aldehydes and ketones can be reduced to alcohols by reacting them with a sodium/ethanol ('alcohol') mixture or a zinc/ethanoic acid mixture, but these reactions are not usually dealt with in UK AS-A2 or IB chemistry. The reduction involves the half-cell reaction: >C=O + 2H⁺ + 2e^- ==> >CHOH. The reaction at the zinc metal surface via the half-cell reaction: Zn_(s) ==> Zn²⁺_(aq) + 2e^-. 8.2(d) I don't know if reduction of aldehydes/ketones with hydrogen/Ni catalyst is used in industry? but it is a feasible and cheap reaction method. aldehyde: RCHO + H₂ ==> RCH₂OH (R = H, alkyl or aryl) ketone: R₂C=O + H₂ ==> R₂CHOH (R = alkyl or aryl) 8.3 Reduction of a carboxylic acid to a primary aliphatic alcohol 8.3(a) NaBH₄, is not a powerful enough reducing agent to reduce carboxylic acids. 8.3(b) LiAlH₄ is a more powerful reducing agent than NaBH₄, and in ether solvent, readily reduces carboxylic acids to primary alcohols. The reaction can be summarised as: RCOOH + 4[H] ==> RCH₂OH + H₂O (R = H, alkyl or aryl) 8.3(c) As far as I know, metal/acid reducing agents are not powerful enough to reduce carboxylic acids. 8.3(d) I don't know if hydrogen/Ni is used to reduce carboxylic acids and I can't see any synthetic use, but it is a feasible reaction. RCOOH + 2H₂ ==> RCH₂OH + H₂O (R = H, alkyl or aryl) 8.4 Reduction of nitriles to primary aliphatic amines 8.4(a) NaBH₄, is not a powerful enough reducing agent to effect the change. 8.4(b) LiAlH₄ is a more powerful reducing agent than NaBH₄ and in ether solvent will reduce nitriles to primary aliphatic amines. The reaction can be summarised as: RCN + 4[H] ==> RCH₂NH₂ (R = H, alkyl or aryl) 8.4(c) Sn/HCl_(aq), is not a powerful enough reducing agent to reduce nitriles to amines. 8.4(d) Hydrogen/150^oC?/Ni catalyst conditions will reduce nitriles to primary aliphatic amines. RCN + 2H₂ ==> RCH₂NH₂ (R = H, alkyl or aryl) 8.5 Reduction of nitro-aromatics to primary aromatic amines 8.5(a) NaBH₄, is not a powerful enough reducing agent to reduce nitro-aromatic compounds. 8.5(b) LiAlH₄ is a more powerful reducing agent than NaBH₄ and in ether solvent readily reduces nitro-aromatics to primary aromatic amines, the simplified equation is ... C₆H₅NO₂ + 6[H] ==> C₆H₅NH₂ + 2H₂O methylnitrobenzenes would be reduced to methylphenylamine primary amines, i.e. CH₃C₆H₄NO₂ + 6[H] ==> CH₃C₆H₄NH₂ + 2H₂O as will any aromatic compound with a nitro group (-NO₂) attached directly to the benzene ring. 8.5(c) The reduction of nitro-aromatics with tin and concentrated hydrochloric acid. In the laboratory, reacting a nitro-aromatic with a mixture of tin and conc. hydrochloric acid by heating under reflux will reduce it to a primary aromatic amine (-NH₂ directly attached to benzene ring). In industry a cheap metal like iron powder and acid are used or a direct reduction in the gas phase with hydrogen/transition metal catalyst (8.5(d). In the 'laboratory' preparation, the mixture may need heating to complete the reaction (can be refluxed?) and formation of phenylamine (aniline) from nitrobenzene can be summarised as C₆H₅NO₂ + 6[H] ==> C₆H₅NH₂ + 2H₂O but the 'real' equations are rather more complicated, the simplest redox equation I can come up with is 2C₆H₅NO_2(aq) + 14H⁺_(aq) + 3Sn_(s) ==> 2C₆H₅NH₃⁺_(aq) + 3Sn⁴⁺_(aq) + 4H₂O_(l) which shows the formation of the phenylammonium cation because the amine is a base and formed in an acid medium. The tin(IV) ion is actually a chlorocomplex ion of tin, SnCl₆^2-, the hexachlorostannate(IV) ion, so the full ionic-redox equation is more correctly written as ... if you really must! 2C₆H₅NO_2(aq) + 14H⁺_(aq) + 18Cl^-_(aq) + 3Sn_(s) ==> 2C₆H₅NH₃⁺_(aq) + 3[SnCl₆]^2-_(aq) + 4H₂O_(l) Oxidation state changes: 3Sn (0) inc. to (+4) balanced by 2N decreasing from (+3) to (-3). and then solid or conc. aqueous sodium hydroxide is added to free the amine (immiscible) from its arylammonium cation C₆H₅NH₃⁺_(aq) + OH^-_(aq) ==> C₆H₅NH_2(l) + H₂O_(l) The primary aromatic amine is then be extracted by steam distillation. For more on the theory of steam distillation see Equilibria Part 8.5 On addition of the strong alkali the amine separates out as an oily layer and the mixture is heated with the steam input. A mixture of the amine and water 'steam distils' into the condenser and separates into two layers in the collection flask which can then be further processed using a separating funnel. The phenylamine may have to be extracted with 'ether' solvent? Phenylamine or mixture with ether, can be dried with solid sodium hydroxide and fractionally distilled to obtain pure phenylamine liquid. 8.5(d) Aromatic nitro-compounds are reduced with hydrogen/Ni or Cu catalyst at elevated temperatures and the resulting primary aromatic amines are very important intermediate compounds in dye and drug manufacture e.g. C₆H₅NO₂ + 3H₂ ==> C₆H₅NH₂ + 2H₂O CH₃C₆H₃(NO₂)₂ + 6H₂ ==> CH₃C₆H₃(NH₂)₂ + 4H₂O

10. Other miscellaneous Organic Redox Reactions

This is a 'collection' of reactions not dealt with in sections 8. and 9. They may/may not be useful reactions.

Section 10. reaction sub-index: 10.1 Cannizzaro reaction * 10.2 aldehydes/ketones tests * 10.3 Combustion * 10.4 Fuel cells * 10.5 The use of 1,4-dihydroxybenzene (quinol, hydroquinol) in photography

10.1 The Cannizzaro reaction
- Aldehydes which do not have a hydrogen atom on the carbon next to the carbon of the carbonyl group (C=O) undergo the Cannizzaro reaction with concentrated aqueous sodium hydroxide in which one molecule of the aldehyde is reduced to a primary alcohol and another is oxidised to the sodium salt of a carboxylic acid.
- This is an organic example of disproportionation in which the same carbon atoms of the reactant molecule simultaneously increase and decrease their oxidation state e.g.
  - methanal changes to methanol and sodium methanoate
    - 2HCHO + Na⁺OH^- ==> CH₃OH + HCOO^-Na⁺
  - benzaldehyde changes to phenylmethanol (benzyl alcohol) and sodium benzoate
    - 2C₆H₅CHO + Na⁺OH^- ==> C₆H₅CH₂OH + C₆H₅COO^-Na⁺
10.2 Simple chemical tests to distinguish aldehydes from ketones
- The tests depend on the relative redox stability of ketones compared to the much more readily oxidised aldehydes. These tests also give positive results with many reducing sugars and some rather more stable aromatic aldehydes e.g. benzaldehyde, may not give a positive result at all.
- In these tests, because aldehydes are stronger reducing agents than ketones, they reduce the metal ion and are oxidised in the process i.e. RCHO + [O] ==> RCOOH, a change which is equivalent to a 2 electron loss by the RCHO.
- Tollens reagent is a colourless solution of silver nitrate in aqueous ammonia.
  - When an aldehyde is warmed with Tollens reagent it is oxidised to a carboxylic acid and the silver ion (in ammine complex form) is reduced to silver, forming a silver mirror on the side of the test tube.
  - 2[Ag(NH₃)₂]⁺_(aq) + R-CHO_(aq) + H₂O_(l) ==> 2Ag_(s) + 4NH_3(aq) + R-COOH_(aq) + 2H⁺_(aq)
  - simplified: 2Ag⁺_(aq) + R-CHO_(aq) + H₂O_(l) ==> 2Ag_(s) + R-COOH_(aq) + 2H⁺_(aq)
  - or 2[Ag(NH₃)₂]⁺_(aq) + R-CHO_(aq) + 2OH^-_(aq) ==> 2Ag_(s) + 4NH_3(aq) + R-COOH_(aq) + H₂O_(l)
  - simplified: 2Ag⁺_(aq) + R-CHO_(aq) + 2OH^-_(aq) ==> 2Ag_(s) + R-COOH_(aq) + H₂O_(l)
  - Ketones show no reaction because of their greater stability to oxidation.
- Fehlings or Benedict's solution consists of a copper(II) ion complexed with an organic carboxylic acid.
  - The deep blue copper(II) ion in the complex is reduced to a red-brown precipitate of copper(I) oxide.
  - Ketones show no reaction due to their greater oxidation stability.
  - 2Cu²⁺_(complex/aq) + R-CHO_(aq) + 2H₂O_(l) ==> Cu₂O_(s) + R-COOH_(aq) + 4H⁺_(aq)
  - or: 2Cu²⁺_(complex/aq) + R-CHO_(aq) + 4OH^-_(aq) ==> Cu₂O_(s) + R-COOH_(aq) + 2H₂O_(l)
  - Ketones show no reaction because of their greater reluctance to oxidation.
- There are lots more organic chemical tests described - Chemical Identification Tests (with alphabetical index).
10.3 All organic compound air/oxygen combustion reactions are oxidations in terms of the carbon atoms of the organic molecule e.g. carbon's oxidation state is increased from (-4) in methane to (+4) in CO₂ for complete combustion and (+2) if carbon monoxide formed or (0) in carbon (soot) if the combustion is inefficient/incomplete. In each case oxygen's oxidation state changes from (0) to (-2) to offset the increase in carbon's oxidation state. e.g.
- CH_4(g) + 2O_2(g) ==> CO_2(g) + 2H₂O_(l)
- 2CH_4(g) + 3O_2(g) ==> 2CO_(g) + 4H₂O_(l)
- CH_4(g) + O_2(g) ==> C_(s) + 2H₂O_(l)
10.4 Organic Fuel Cells
- Many hydrocarbon molecules are burned as fuels, but since it is a redox reaction, theoretically it can be done as a combined half-cell oxidation/reduction electron transfer reaction.
- In practice organic compounds can be oxidised by oxygen in a way that can be used to generate electricity directly in a fuel cell (right diagram) rather than release the energy as heat e.g.
- In reaction 9.1(a) ethanol was oxidised to ethanoic acid by acidified potassium dichromate(VI). The same result can be obtained by reaction of ethanol with oxygen in a fuel cell.
- The reaction is very exothermic, so ethanol is a good source of chemical potential energy.
- Ethanol is used directly as a fuel in a direct ethanol fuel cell (a DEFC fuel cell).
- (i) CH₃CH₂OH_(aq) + O_2(g) ==> CH₃COOH_(aq) + H₂O_(l) (ΔH^Ø = -494 kJ mol^-1)
- The half-cell reactions are:
  - (ii) Reduction, +ve electrode: O_2(g) + 4H⁺_(aq) + 4e^- ==> 2H₂O_(l) (E^Ø = +1.23V)
  - (iii) Ox'n, -ve electrode: CH₃CH₂OH_(l) + H₂O_(l) ==> CH₃COOH_(aq) + 4H⁺_(aq) + 4e^- (E^Ø = +0.06V)
  - Adding (ii) + (iii) = equation (i) and E^Ø_cell = E^Ø_+/red'n - E^Ø_-/ox'n = 1.23 - 0.06 = +1.17V
  - The reactions must take place on catalytic electrodes made of platinum and other transition metals and here the inner 'electrolyte' is as a polymer proton exchange membrane of the fuel cell (a PEFC fuel cell), though it can be a concentrated phosphoric acid solution (in a PAFC fuel cell).
  - The electrons will flow through the external circuit from the -ve electrode to the +ve electrode.
  - Free energy change: ΔG^Ø = -nE^ØF = - 4 x 1.17 x 96500 = -451620 J mol^-1 = -451.6 kJ mol^-1
  - n = number of electrons transferred, E^Ø = cell voltage, F = Faraday constant in coulombs mol^-1
  - This sort of chemistry is being developed to make reasonably efficient portable fuel cells
- Commercially developed fuel cells will hopefully convert the ethanol completely into water and carbon dioxide to give a greater and more efficient energy output, but its still generates a voltage of 1.0-1.2V per cell, in this case ...
  - (i) Oxidation, -ve electrode: C₂H₅OH_(l) + 3H₂O_(l) ==> 12H⁺_(aq) + 2CO_2(aq/g) + 12e^-
  - (iii) Reduction, +ve electrode: 3O_2(aq/g) + 12H⁺_(aq) + 12e^- ==> 6H₂O_(l)
  - adding (i) + (iii) gives: C₂H₅OH_(l) + 3O_2(aq/g) ==> 3H₂O_(l) + 2CO_2(aq/g)
  - + other organic products as the reaction is not completely efficient.
  - The cells can obviously be connected in series to give larger voltages.
  - The ethanol ('alcohol', C₂H₅OH) can be bio-sourced from sugar beet, potatoes and cereal crops and other plant material that can fermented with enzymes. The ethanol is fractionally distilled from the fermented mixture and constitutes a renewable fuel.
- However there seems to have been more work done? on the Direct Methanol Fuel Cell (a DMFC fuel cell), though there are concerns over methanol's toxicity and the very costly platinum catalytic electrodes required, but the DMFC is essentially like the DEFC described above and the principles illustrated in the diagram.
- The methanol can be synthesised in the reforming reaction CO_(g) + 2H_2(g) ==> CH₃OH_(l)
  - half-cell reactions for the DMFC cell:
    - (i) Red'n, +ve electrode: ³/₂O_2(g) + 6H⁺_(aq) + 6e^- ==> 3H₂O_(l) (E^Ø = +1.23V)
    - (ii) Ox'n, -ve electrode: CH₃OH_(aq) + H₂O_(l) ==> CO_2(aq/g) + 6H⁺_(aq) + 6e^- (E^Ø = ?V)
    - so overall the cell reaction is ... with a maximum output voltage of about 1.0V.
    - CH₃OH_(aq) + ³/₂O_2(g) ==> CO_2(aq/g) + 3H₂O_(l)
- Hopefully, efficient cells using ethanol can be developed because ethanol compared to methanol has a higher energy density (e.g. kJ/kg), is less toxic and can be bio-resourced as a renewable fuel.
- There is a huge amount of research going on in fuel cell development to try to use cheaper, but equally effective tiny particle metal catalysts of Fe, Co and Ni instead of costly platinum, but the most efficient metals are still the most costly?
10.5 The use of 1,4-dihydroxybenzene (p-quinol, hydroquinol) in photography
- 1,4-dihydroxybenzene (Quinol) is an ingredient in photographic developing solutions. In the transparent plastic emulsion film of the exposed film it will reduce the silver ions in silver halide salts, (which have not already been decomposed by light to silver and bromine), to silver as well. It would appear that these extra silver atoms produced by the developer, cluster around the already formed silver atoms (acting as 'nuclei') from the action of light, to enhance the image in 'shades' and 'contrast' to give a more clearly defined negative.
  - The relevant half-cell reactions and their standard redox electrode potentials are ...
  - AgBr_(s) + e^- ==> Ag_(s) + Br^-_(aq) (E^θ_Ag/AgBr = +0.073V at 298K)
  - _(aq) + 2H⁺_(aq) + 2e^- ==> _(aq) (E^θ_Q/QH2 = +0.059V at 298K, pH 8.5)
  - These two organic molecules are referred to by a variety of systematic and trivial names, confusing when searching the web! e.g.
    - left (ref Q): 2,5-cyclohexadiene-1,4-dione, p-quinone, ('one' higher ranking than 'ene')
    - right (ref QH2): 1,4-dihydroxybenzene, 1,4-benzenediol, benzene-1,4-diol, p-quinol, p-hydroquinone, p-dihydroxybenzene
- It is convention to show the half-cell reactions as reductions when quoting them with the half-cell potential.
- In the reactions the silver ions in the silver bromide salt are reduced to silver atoms and 1,4-dihydroxybenzene (quinol) is oxidised to 2,5-cyclohexadiene-1,4-dione (quinone). The bromide ions 'dissolve' and two hydrogen ions are released from the quinol. Therefore the basic reaction via the alkaline developer is ....
  - _(aq) + 2AgBr_(s) ==> _(aq) + 2Ag_(s) + 2Br^-_(aq) + 2H⁺_(aq)
  - Sometimes more simply written as ...
  - C₆H₄(OH)_2(aq) + 2AgBr_(s) ==> C₆H₄O_2(aq) + 2Ag_(s) + 2HBr_(aq)
  - E^θ_reaction = E^θ_red - E^θ_ox = E^θ_AgBr/Ag - E^θ_Q/QH2
  - E^θ_reaction = (+0.073) - (+0.059) = +0.014V
  - The positive value for E^θ_reaction shows the reaction is feasible. In fact the E^θ_Q/QH2 becomes less positive as the pH is increased making the reaction more feasible, but the E_AgBr/Ag is theoretically independent of pH. The hydrogen ions would be neutralised by the alkaline media.
- Oxidation number analysis for the molecules in question is shown below ...
  - using graphic structural formulae and
  - Oxidation states in organic compounds is explained in section 11. below.
- Note: Not all the silver bromide reacts in this way, so in the fixing process the remaining silver bromide is removed by sodium thiosulphate to produce the 'light stable' negative.
- The transition metal complex ion [Ag(S₂O₃)₂]^3-_(aq) is formed when sodium thiosulphate (Na₂S₂O₃) is used to remove unreacted silver bromide (AgBr) crystals in developing photographic films.
  - AgBr_(s) + 2S₂O₃^2-_(aq) ==> [Ag(S₂O₃)₂]^3-_(aq) + Br^-_(aq)
  - This is NOT a redox reaction, Ag is +1 and Br is -1 throughout the reaction. The thiosulfate ion is here acting as a ligand and not a reducing agent e.g. like its reaction with iodine.
-

11. Oxidation state and organic compounds

Usually the oxidation state of hydrogen is +1, and oxygen -2 in organic compounds.

(The quoted Pauling electronegativities are C = 2.5, H = 2.2 and O = 3.5 which gives the lead in assigning oxidation numbers in this section i.e. the highest oxidation state is assigned to the least electronegative atom and vice versa for hydrogen and oxygen BUT beware for carbon, logical deduction can give some surprising, but correct results!)

On this basis you can achieve a useful oxidation number analysis of simple organic compounds in an oxidation sequence.

e.g. the oxidation sequence below, with the oxidation state of carbon in () and in hydrogen in ().

CH₄ (-4) == ox'n ==> CH₃OH (-2) == ox'n ==> HCHO (0) == ox'n ==> HCOOH (+2) ==> CO₂ (+4)

The above sequence can described in terms of the 'level of the functional group' which is equal the number of bonds the carbon atom forms with more electronegative atoms like oxygen. Therefore in the above sequence: alkane hydrocarbons are level zero (carbon and hydrogen have virtually the same electronegativity), alcohols are level 1 (C-OH, as are halogenoalkanes/haloalkanes e.g. C-Cl), carbonyl is level 2 (e.g. C=O in aldehydes/ketones), carboxylic acids are level 3 (C=O and C-OH) and finally the fully oxidised carbon in carbon dioxide is level 4. Note that a functional group level applies to a single carbon atom e.g. the carbon of the functional group, and not the full molecule.

Similarly for the oxidation sequence from ethane to ethanoic acid ...

CH₃CH₃ (-3,-3) = ox'n => CH₃CH₂OH (-3,-1) = ox'n => CH₃CHO (-3,+1) = ox'n => CH₃COOH (-3,+3)

Note the rise of carbon's oxidation state in increments of 2, see oxidation equations for acidified potassium dichromate(VI) reaction with alcohols and aldehydes in section 9.1(a) where the half-cell oxidation equations involve a 2 electron loss from the organic molecule.

Other organic molecules and redox sequences can be similarly 'analysed'

ethene H₂C=CH₂ (-2,-2) + H₂ (0) == reduction/Ni ==> ethane CH₃-CH₃ (-3,-3), (+1)

propene CH₃-CH=CH₂ (-3,-1,-2) + H₂ == reduction/Ni ==> CH₃-CH₂-CH₃ (-3,-2,-3)

ethanol CH₃-CH₂-OH (-3,-1) == ox'n ==> ethanal CH₃CHO (-3,+1) == ox'n ==> CH₃COOH (-3,+3)

Appendix 1. The Concept of Functional Group Level

This concept of functional group level has been introduced into some UK A level pre-university courses.
It can be related to an increasingly oxidation state of carbon.
In the explanation with examples below assume R = a H, alkyl or aryl grouping.
The concept of functional group level of a carbon atom is derived from counting the number of bonds form an individual carbon atom to electronegative atoms (C-X, where I assume X more electronegative than carbon i.e. oxygen, nitrogen or halogens such as chlorine).
If no such C-X bonds exist the carbon is described as being at the hydrocarbon level.
With one C-X bond the carbon atom is at the alcohol level e.g. alcohols (C-OH), monohaloalkanes (R₃C-Cl).
With two C-X bonds the carbon atom is at the carbonyl level e.g. aldehydes and ketones (R₂C=O), ethers (R₃C-O-CR₃)
With three C-X bonds the carbon atom is at the carboxylic acid level RCOOH (RO-C=O grouping), esters RCOOR, amides RCONH2, nitrile RCN (RN)
Finally with four C-X bonds we reach the carbon dioxide level e.g. carbon dioxide itself CO₂ or an ether such as C(OR)₄
The concept of functional group level often applies to a single carbon atom, not the whole molecule e.g.
- alcohol COH, chloroalkane C-Cl, aldehyde CHO,
- carboxylic acid level: carboxylic acid COOH, amide CONH₂,
- but not exclusively e.g. C=C in alkenes, but would this would be considered at the hydrocarbon level despite being at a higher oxidation state than a saturated alkane? (see section 11).
If a reaction takes place within a level you are swapping one heteroatom (non-carbon atom) for another
- e.g. hydrolysis of a halogenoalkane: R₃C-Cl + NaOH ==> R₃COH + NaCl
- which is occurring at the alcohol level
In order to move a carbon atom up a level requires an oxidizing agent (dealt with on this page in section 9).
- e.g. the oxidation of alcohols using acidified potassium dichromate(VI) solution
- R-CH2OH == Cr₂O₇^2-/H⁺ ==> RCHO == Cr₂O₇^2-/H⁺ ==> RCOOH
- alcohol level ==> carbonyl level ==> carboxylic acid level
To move a carbon atom down a level requires a reducing agent (dealt with in section 8) or carbanion equivalent (not dealt with yet)
- e.g. the reduction of carboxylic acids to primary alcohols with lithium tetrahydridoaluminate(III)
- RCOOH + 4[H] ==> RCH₂OH + H₂O
- In this case the carbon atom of the functional group is reduced two levels i.e. from the carboxylic level to the alcohol level.
-

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(italian) Doc Brown chimica REDOX CHIMICA Parte 3 Le sezioni 8-10 reazioni di ossidazione e riduzione Preparativa in C hemistry Organica sezione redox- index: 1. definizioni redox di base * 2. Introduzione stato di ossidazione (con il sub-indice) * 3. ossidazione norme dello Stato-le linee guida ed esempi inorganici *4. Naming composti inorganici * 5. Utilizzo di stati di ossidazione per descrivere i cambiamenti redox in uno inorganico equazione di reazione dato (con il sub- indice) * 6. Costruire la piena equazioni redox inorganica da mezza equazioni (con il sub-indice) * 7. titolazioni redox * 8. riduzioni di organici di sintesi (con tabella riassuntiva) * 9. Sintesi organica ossidazioni (con tabella riassuntiva) * 10. organici Altre reazioni redox (con il sub-indice) * 11. bue di carbonio. state in selected organic compounds, functional group level 1. stato scelto in composti organici, funzionali a livello di gruppo * Si consiglia di studiare le sezioni in rigoroso ordine e copre le esigenze di-A2 * AS Vedi anche Equilibri Parte 7 reazioni redox per Half equilibri cellulari, elettrodo standard di idrogeno elettrodo, il potenziale, le cellule semplici e notazione, Serie elettrochimica, E cella Ø fattibilità per reazione, 'batterie' e sistemi di celle a combustibile, ecc * (portugueses) Doc Brown Química REDOX QUÍMICA Parte 3 Secções 8-10 de oxidação e redução Preparative reações orgânicas reações redox em Organic Chemistry Um resumo de algumas reações redox utilizados em síntese orgânica é dado abaixo. Detalhes adicionais sobre reações selecionadas são indicadas a seguir as tabelas de resumo. A maioria das reações descritas são encontradas em uma ou outra baseada no Reino Unido ou cursos. A aplicação dos estados de oxidação de moléculas orgânicas pode ser complicado, mas, (i) a utilização de células-metade equações normalmente fica em torno do problema, e (ii) espero que o estado de oxidação exemplares na última seção ajudará a esclarecer a situação, se você está interessado, mas esse conhecimento não é necessária a este nível? Importante: O exame menos oxigênio ou gás hidrogênio é usado diretamente na síntese de reação redox [O] e [H] deve ser utilizado em equações simplificadas e exemplos serão citados em cada seção e algumas grades curriculares especificamente estado assim. O conceito de nível de grupo funcional é abordada no apêndice 1. * (spanish) Doc Brown Química REDOX QUÍMICA Parte 3 Secciones 8 a 10 de oxidación y reducción en las reacciones preparativas Orgánica 8. reducciones de la síntesis orgánica (con cuadro resumen) * 9. Síntesis orgánica oxidaciones (con cuadro resumen) * 10. Orgánica Otras reacciones redox (con subíndice) * 11. buey de carbono. state in selected organic compounds, functional group level * You are advised to study sections 1. Estado en determinados compuestos orgánicos, el grupo de nivel funcional * (thai) Brown 's เคมี Doc รีดอกซ์ ส่วนเคมี 3 ส่วน 8-10 ออกซิเดชันและลดการตอบโต้เตรียมใน C hemistry อินทรีย์ ปฏิกิริยารีดอกซ์ใน สังเคราะห์สรุปปฏิกิริยาของปฏิกิริยาที่ใช้อินทรีย์ได้รับด้าน ล่าง รายละเอียดเพิ่มเติมสำหรับปฏิกิริยาที่เลือกจะรับด้านล่างตารางสรุป ที่สุดของปฏิกิริยาที่อธิบายที่พบในหนึ่งหรืออื่น ๆ ของสหราชอาณาจักรตาม GCE - AS - A2 หรือหลักสูตร IB . ใบสมัครของรัฐออกซิเดชันที่ โมเลกุลอินทรีย์สามารถหากิน แต่, (i) การใช้ - สมเซลล์จะได้รับครึ่งรอบ ปัญหาและ (ii) หวัง exemplars เลขออกซิเดชันในส่วนสุดท้ายจะช่วยให้แสงสว่างสนใจสถานการณ์หากคุณ แต่ความรู้นี้ไม่จำเป็นต้องอยู่ใน ระดับนี้. บันทึกการตรวจสอบข้อสำคัญ : หากหรือก๊าซออกซิเจนก๊าซไฮโดรเจนใช้โดยตรงในการสังเคราะห์ปฏิกิริยารีดอกซ์ [O] และ [H] ควรจะใช้ในสมการง่ายและจะยกตัวอย่างในแต่ละส่วนและ syllabuses เฉพาะบางรัฐเพื่อ 8 อินทรีย์ลดการสังเคราะห์ (พร้อมตารางสรุป) * 9 สังเคราะห์อินทรีย์ oxidations (พร้อมตารางสรุป) * 10 อื่น ๆ อินทรีย์ปฏิกิริยาตอบโต้ (with sub - index) * 11 ของวัว รัฐในสารอินทรีย์ เลือกกลุ่มทำงานระดับ * คุณควรศึกษา 1 ส่วน t * (russian) Doc Брауна химии Редокс ХИМИЯ Часть 3 Разделы 8 до 10 Окисление и сокращении Препаративная реакции в органических 8. Органический синтез сокращение (с сводная таблица) * 9. Органический синтез окисления (с сводная таблица) * 10. других органических окислительно-восстановительных реакций (с суб-индекс) * 11. углерода в быка. state in selected organic compounds, functional group level * You are advised to study sections 1. состояние в отдельных органических соединений, функциональном уровне группы * Вы посоветовали Окислительно- восстановительных реакций в органической химии Обзор некоторых окислительно-восстановительных реакций в органическом синтезе, приводится ниже. Более подробная информация по отдельным реакций приведены ниже сводные таблицы. Большинство описанных реакций находятся в той или иной британской GCE-AS-A2 или IB курсов. Применении окисления органических молекул может быть сложно, но, (я) использование полуэлементом уравнений обычно обходит проблемы, и (II), мы надеемся, окисления экземпляров в последнем разделе помогут осветить ситуацию, если вы хотите, но это знание не требуется на этом уровне? Важное исследование к сведению: Если водород или кислород используется непосредственно в окислительно-восстановительные реакции синтеза [O] и [H] должны использоваться в упрощенных уравнений и примеров, будут указаны в каждом разделе, а некоторые учебные конкретно указывается, так. *

9. Summary of some ORGANIC SYNTHESIS OXIDATION REACTIONS Guide notes Some direct catalytic oxidations at higher temperatures used in industry are included, but most are 'school laboratory' reactions. YES/NO - for a laboratory synthesis. R = alkyl or aryl for primary/secondary alcohols. Click on 9.1, 9.2, 9.3, 9.4 and 9.5 in the table for more details on (a), (b) etc. Important examination note: Unless hydrogen gas or oxygen gas is used directly in the redox synthesis reaction [O] and [H] should be used in simplified equations and examples will be quoted in each section and some syllabuses specifically state so.
homologous series change on oxidation	molecular structure change	(a) heat with mod conc. H₂SO₄ and K₂Cr₂O_7(aq) (lab method)	(b) reflux with KMnO₄/NaOH_(aq) (lab method)	(c) oxygen + catalyst or thermal decomposition (industrial methods)
9.1 primary alcohol ==> aldehyde ==> carboxylic acid	RCH₂OH ==> RCHO ==> RCOOH	YES	YES but only get RCOOH and of little synthetic use	e.g. CH₃CH₂OH ==> CH₃CHO (Cu/500^oC)
9.2 secondary alcohol ==> ketone	R₂CHOH ==> R₂C=O	YES	YES but of little synthetic use	(CH₃)₂CHOH ==> CH₃COCH₃ (Cu/500^oC)
9.3 tertiary alcohol ==> ?	R₃C-OH fairly stable (if oxidised C-C bonds broken ==> lower RCOOH, CO₂, H₂O)	not readily oxidised - no synthetic use	not readily oxidised - no synthetic use	not readily oxidised - no synthetic use
9.4 alkyl groups on benzene ring	e.g. C₆H₅CH₃ ==> C₆H₅COOH	YES	YES	YES air/150^oC/Co salt
9.5 alkene ==> ?	details in appropriate box	NO	ethene ==> ethane-1,2-diol (at room temp.)	e.g. ethene ==> epoxyethane (Ag/250^oC)
Some further details of the organic oxidations tabulated above 9.1 Oxidation of primary alcohols to aldehydes and carboxylic acids (includes oxidation of aldehydes to carboxylic acids) 9.1(a) It is possible using the same reagent of aqueous sodium/potassium dichromate(VI)-sulphuric acid to oxidise a primary alcohol to either the aldehyde, or the carboxylic acid, depending on the reaction conditions. In order to selectively isolate the aldehyde this initial oxidation products must be removed from the reaction mixture as quickly as possible, otherwise oxidation proceeds to the carboxylic acid. The method involves heating under reflux if an aldehyde/ketone is to be prepared in the way illustrated in the detailed diagram PD1 below. The 25% sulphuric acid is placed in the flask and gently simmered. The alcohol and aqueous sodium/potassium dichromate(VI) solution is dripped onto the hot acid. Immediately, the orange dichromate(VI) is reduced by the alcohol to the green chromium(III) ion and the alcohol is oxides to the aldehyde or ketone. The technique illustrated above is called heating under reflux, a method which enables a reaction to be carried out at a higher temperature than room temperature to speed up the reaction AND retain the solvent (reaction medium e.g. water) and any volatile reactant or product (e.g. an alcohol/aldehyde/ketone). As the mixture boils, the vapours of the solvent or volatile reactant/product are condensed back into the flask in the vertical condenser, so any volatile reactant is used up and no volatile product lost (at least at this stage in a preparation!). The diagram shows a bunsen burner being used to supply the heat ('my days'), these days its more likely, and safer, to use an electrical heater that the round bottomed flask fits in snugly. A spot of theory to explain the separation of the aldehyde/ketone from the reaction mixture. For the same carbon number, the boiling point of the polar aldehyde/ketone (^δ+C=O^δ-, but no H bonding) is lower than the original more polar alcohol (^δ-O-H^δ+, hydrogen bonding) whose bpt. is higher. Therefore, as long as the bpt. of the aldehyde/ketone is not too high, in the set-up shown above, the aldehyde rapidly distils over and condenses in the collection tube/flask with some water. This rapid in situ extraction ensures that most of the aldehyde (or ketone), 9.1(a) is not oxidised further. If the carboxylic acid of the same carbon number is required from a primary alcohol, the mixture is refluxed using the set-up illustrated in diagram PD2. (i) primary alcohol ==> aldehyde Cr₂O₇^2-_(aq) + 3RCH₂OH_(aq) + 8H⁺_(aq) ==> 3RCHO_(aq) + 2Cr³⁺_(aq) + 7H₂O_(l) reduction half reaction: Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6e^- ==> 2Cr³⁺_(aq) + 7H₂O_(l) oxidation half reaction: RCH₂OH_(aq) ==> RCHO_(aq) + 2H⁺_(aq) + 2e^-_(aq) (R = alkyl or aryl) Examples using simplified symbol equations: ethanol ==> ethanal: CH₃CH₂OH + [O] ==> CH₃CHO + H₂O propan-1-ol (1-propanol, n-propyl alcohol, n-propanol) ==> propanal CH₃CH₂CH₂OH + [O] ==> CH₃CH₂CHO + H₂O then under reflux conditions the further oxidation ... (ii) aldehyde ==> carboxylic acid Cr₂O₇^2-_(aq) + 3RCHO_(aq) + 8H⁺_(aq) ==> 3RCOOH_(aq) + 2Cr³⁺_(aq) + 4H₂O_(l) (R = alkyl or aryl) oxidation half-reaction: RCHO_(aq) + H₂O_(l) ==> RCOOH_(aq) + 2H⁺_(aq) + 2e^-_(aq) Examples using simplified symbol equations: ethanal ==> ethanoic acid, CH₃CHO + [O] ==> CH₃COOH propanal (propionaldehyde) ==> propanoic acid (propionic acid) CH₃CH₂CHO + [O] ==> CH₃CH₂COOH butanal (butyraldehyde) ==> butanoic acid (butyric acid) CH₃CH₂CH₂CHO + [O] ==> CH₃CH₂CH₂COOH so overall for reflux conditions (i) + (ii) gives (iii) primary alcohol ==> carboxylic acid 2Cr₂O₇^2-_(aq) + 3RCH₂OH_(aq) + 16H⁺_(aq) ==> 3RCOOH_(aq) + 4Cr³⁺_(aq) + 11H₂O_(l) oxi'n half-reaction: RCH₂OH_(aq) + H₂O_(l) ==> RCOOH_(aq) + 4H⁺_(aq) + 4e^-_(aq) (R = alkyl or aryl) Examples using simplified symbol equations: ethanol ==> ethanoic acid CH₃CH₂OH + 2[O] ==> CH₃COOH + H₂O propan-1-ol (1-propanol, n-propyl alcohol, n-propanol) ==> propanoic acid CH₃CH₂CH₂OH + 2[O] ==> CH₃CH₂COOH + H₂O 9.1(b) Heating a primary alcohol with a aqueous sodium hydroxide/potassium manganate(VII) mixture under reflux (diagram PD2) will give the sodium salt of the carboxylic acid and it is not possible to isolate the intermediate aldehyde. However, the acid/dichromate(VI) method 9.1(a) under reflux is better, and the carboxylic acid is less liable to further degradative oxidation. The complex reaction can be summarised as: RCH₂OH_(aq) + NaOH_(aq) + 2[O] ==> RCOO^-Na⁺_(aq) + 2H₂O_(l) (R = alkyl or aryl) After removing the excess KMnO₄/MnO₂ the weak acid is freed from its sodium salt by adding strong dilute hydrochloric acid. RCOO^-_(aq) + H⁺_(l) ==> RCOOH 9.1(c) Many chemical feedstocks are oxidised directly with molecular oxygen/transition metal catalyst to produce useful products in industry. e.g. 2CH₃OH + O₂ ==> 2HCHO + 2H₂O (Ag/500^oC, methanol ==> methanal) or 2CH₃CH₂OH + O₂ ==> 2CH₃CHO + 2H₂O (Ag/500^oC, ethanol ==> ethanal) and the latter reaction can also be achieved via a thermal decomposition using a different catalyst, e.g. CH₃CH₂OH ==> CH₃CHO + H₂ (Cu/500^oC), which is still an oxidation, right carbon (-1) to (+1) and 2 x hydrogen (+1) to (0). 9.2 Oxidation of secondary alcohols to ketones 9.2(a) In the case of secondary alcohols you only get the ketone, unless you reflux the alcohol/K₂Cr₂O₇/H₂SO_4(aq) for a long time, in which case the ketone can be oxidised to lower carbon number carboxylic acids, carbon dioxide and water etc. if the carbon chain is broken Ketones are quite stable to further oxidation due to the strong carbon-carbon (C-C) bonds that have to be broken. To be on the safe side it is better to make the ketone under the same restricted reaction conditions used to produce the aldehyde (details above with diagram PD1). Cr₂O₇^2-_(aq) + 3R₂CHOH_(aq) + 8H⁺_(aq) ==> 3R₂C=O_(aq) + 2Cr³⁺_(aq) + 7H₂O_(l) oxidation half-reaction: R₂CHOH_(aq) ==> R₂C=O_(aq) + 2H⁺_(aq) + 2e^-_(aq) (R = alkyl or aryl) 9.2(b) Ketones are produced by refluxing secondary alcohols with NaOH/KMnO_4(aq), but further oxidation is likely to take place because this reagent is a stronger oxidising agent than acidified potassium dichromate(VI). (CH₃)₂CHOH + [O] ==> (CH₃)₂C=O + H₂O (propan-2-ol ==> propanone) 9.2(c) Many chemical feedstocks are oxidised directly with molecular oxygen/transition metal catalyst to produce useful products in industry. 2(CH₃)₂CHOH_(g) + O_2(g) ==> 2(CH₃)₂C=O_(g) + 2H₂O_(g) (Ag/500^oC) and this reaction can also be achieved via a thermal decomposition using a different catalyst. (CH₃)₂CHOH_(g) ==> (CH₃)₂C=O_(g) + H_2(g) (Cu/500^oC) which is still an oxidation, right carbon (-1) to (+1) and 2 x hydrogen (+1) to (0). 9.3 Oxidation of tertiary alcohols 9.3(a)-(d) Tertiary alcohols, R₃COH (R = alkyl or aryl), are not readily oxidised because strong carbon-carbon bonds have to be broken. The products would be lower chain carboxylic acids, carbon dioxide and water and therefore of no synthetic use. 9.4 Oxidation of alkyl-aromatic hydrocarbons to aromatic carboxylic acids 9.4(a) Acidified potassium dichromate(VI) will oxidise alkyl benzene compounds to benzoic acid, but I think it is slower than with the alkaline manganate(VII) method described below. Overall change is represented by the equations ... C₆H₅CH₃ + 3[O] ==> C₆H₅COOH + H₂O 9.4(b) Aromatic are not easily oxidised and longish reflux times are necessary (illustrated, fig. PD2 below). Hydrocarbons are difficult to oxidise with typical organic oxidising agents compared to compounds like alcohols. However, aromatic hydrocarbons with an alkyl side chain can be oxidised with strong reagents such as aqueous potassium manganate(VII)/sodium hydroxide. Whatever the length of the alkyl group on a benzene ring it gets whittled down to carbon of the carboxylic acid group e.g. propyl benzene ends up as benzoic acid. The more stable aromatic benzene ring is left intact. The overall process for producing benzoic acid from methylbenzene can be summarised .. C₆H₅CH₃ + NaOH + 3[O] ==> C₆H₅COO^-Na⁺ + 2H₂O After removing the excess KMnO₄/MnO₂ the weak benzoic acid is freed from its sodium salt by adding strong dilute hydrochloric acid. C₆H₅COO^-_(aq) + H⁺_(l) ==> C₆H₅COOH or in principle you eventually get benzene-1,2-dicarboxylic acid from 1,2-dimethylbenzene + 6[O] ==> + 2H₂O 9.4(c) Aromatic hydrocarbons with alkyl groups can be directly oxidised with air/oxygen at elevated temperatures and pressures. C₆H₅CH_3(g) + ³/₂O_2(g) ==> C₆H₅COOH_(g) + H₂O_(g) (e.g. air/150^oC/Co salt catalyst) 9.5 Oxidation of alkenes 9.5(a) There are no useful oxidations of alkenes with acidified potassium dichromate(VI) as far as I know. 9.5(b) At room temperature alkenes react with alkaline potassium manganate(VII), KMnO₄/NaOH_(aq), to form diols e.g. ethene ==> ethane-1,2-diol (at room temp.) CH₂=CH₂ + H₂O + [O] ==> 9.5(c) In industry alkenes can be oxidised directly with molecular oxygen to useful products e.g. epoxyalkanes at elevated temperatures and transition metal catalysts. ethene + oxygen ==> epoxyethane (Ag catalyst/250^oC) 2CH₂=CH₂ + O₂ ==> 2 propene + oxygen ==> epoxypropane (Ag catalyst/250^oC ?) 2CH₃-CH=CH₂ + O₂ ==> 2